$h(x)=x^2-7x-8$ 1) What are the zeros of the function? Write the smaller $x$ first, and the larger $x$ second. $\text{smaller }x=$
Solution: To find the zeros of the function, we need to solve the equation $h(x)=0$. We can do that by factoring $h(x)$. $\begin{aligned} x^2-7x-8&=0 \\\\ (x-8)(x+1)&=0 \\\\ x-8=0&\text{ or }x+1=0 \\\\ x={8}&\text{ or }x={-1} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $x$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }x\text{-coordinate}&=\dfrac{({8})+({-1})}{2} \\\\ &={\dfrac72} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $h\left({\dfrac72}\right)$ : $\begin{aligned} h\left({\dfrac72}\right)&=\left({\dfrac72}\right)^2-7\left({\dfrac72}\right)-8 \\\\ &=\dfrac{49}{4}-\dfrac{49}{2}-8 \\\\ &=-\dfrac{81}{4} \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }x&=-1 \\\\ \text{larger }x&=8 \end{aligned}$ The vertex of the parabola is at $\left(\dfrac72,-\dfrac{81}{4}\right)$